Find all positive integers $k,m,n$ satisfying: $\frac1k+\frac1m+\frac1n=\frac{1}{1996}$
The trivial answer is: $k=m=n=3*1996$
$kmn=1996(km+mn+nk)=499\times4\times(km+mn+nk)$ , now $kmn$ must be divisible by all factors in RHS , but how can we get more info about $k,m,n$ ?
Solutions only for $k=m$.
Assume $k=m$, then $\frac{1}{1996}=\frac{2n+m}{mn}$ or $mn=1996\cdot 2n + 1996m$ or $$mn - 1996\cdot 2n - 1996\cdot m =0$$
or $$(n-1996)(m-1996\cdot 2)=2\cdot 1996^2$$
or:
$$(n-1996)(m-1996\cdot 2)=2\cdot 1996^2.$$
If $d\mid 2\cdot 1996^2$ then $n=1996+d$ and $m-1996\cdot 2 = \frac{2\cdot 1996^2}{d}$ or $m=1996\cdot 2 + \frac{2\cdot 1996^2}{d}$.
Since $1996=2^2\cdot 499$, then $d\mid 2^5\cdot 499^2$, which yields already 18 possible solutions.
The extreme case, when $d=1$, gives $n=1997$ and $m=k=2\cdot 1996 \cdot 1997$. When $d=2\cdot 1996^2$, $n=1996(1+2\cdot 1996)$ and $m=k=1996\cdot 2 + 1$.
Algorithm for finding all soluions
If you assume $k<m,n$, then $1996<k<3\cdot 1996$, and you want:
$$\frac{k-1996}{1996k}=\frac{1}m+\frac1n$$
Or $$(k-1996)^2mn - 1996k(k-1996)(m+n)$$
or $$\left((k-1996)m-1996k\right)\left((k-1996)n-1996k\right)=1996^2k^2$$
So you need to find a divisor $d\mid 1996^2k^2$ where both $k-1996\mid d+1996k$ and $k-1996\mid \frac{1996^2k^2}{d}+1996k$, or if:
$$k-1996\mid d+1996^2\\k-1996\mid \frac{1996^2k^2}{d}+1996^2$$
We could write a computer program to enumerate such pairs $(k,d)$ pretty easily, and then solve for $m,n$.
My ruby script turned up 1517 answers, assuming $k\leq m\leq n$, which turns out to be too large for Stack Exchange to handle as raw data, but here is the complete list of $(k,m)$ pairs.